Is this statement above even true? (I think so as I have a proof, but this doesn't have to mean anything)

It appears to me that the latter part of this characterization is a quite strong assumption as $\mathcal{A}$ might contain a lot of additional information, so is there a possibility to weaken it? Or could you mention any similar statements to the one above?

Thank you

EDIT: I accepted the answer of Philip, simply because he has lower points. Francois answer would have deserved it too.

2 Answers
2

(I first wanted to give an answer, but I was not quick enough. I then wanted to add a small comment and found out after 20 minutes that I had insufficient reputation.)

The comment was regarding 2) of oktan's original query: having $H(\theta)$
in the structure is overkill: it suffices to have $( \kappa, <, \in, C)$. (One does not need the structure to be able to express $C$ is closed.'')

Philip, you now have sufficient rep now to comment. But isn't "C is closed" expressible in your structure?
–
Joel David HamkinsJun 22 '10 at 18:33

Thanks Joel. Your are right about the expressibility. My telegraphic, and poorly put, comment was only that one does not {\em need} that "C is closed" to be internally expressible (which came up in the previous answer). Unboundedness suffices.
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Philip WelchJun 22 '10 at 19:50

The forward direction is clear since the set
$$C_{\mathcal{A}} = \{\sup(M\cap\kappa) : M \prec \mathcal{A}, |M| < \kappa\}$$
is a club. Indeed, let $\langle M_\alpha : \alpha < \kappa \rangle$ be an elementary chain of elementary submbodels of $\mathcal{A}$ with size less than $\kappa$ such that:

For the converse, let $C \subseteq \kappa$ be a closed unbounded set and consider the structure $\mathcal{A} = (H(\theta),{\in},{<},C)$. If $M \prec \mathcal{A}$ then $M$ satisfies "$C$ is closed unbounded in $\kappa = \sup C$," and so $C \cap M$ is closed unbounded in $\sup(C \cap M) = \sup(\kappa \cap M)$. It follows that $C_{\mathcal{A}} \subseteq C$, where $C_{\mathcal{A}}$ is defined as above. Thus it is sufficient to consider the structures $\mathcal{A}$ as I just described.